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I'm given an ODE $$y' = \frac{x + 2y + 1}{2x + 3},~y(5/2) = 1/4.$$ To solve this I'm doing the following:

  1. Solve system of equations $$2x_0 + 2y_0 = 1,$$ $$2x_0 = 3,$$ this gives me solutions $x_0 = \frac{3}{2}$, $y_0 = -\frac{1}{4}$. Then, I'm transforming variables $s = x + \frac{3}{2}$, $t = y - \frac{1}{4}$.
  2. Transform the ODE using those new variables to the form $$t' = \frac{s + 2t}{2s}$$ and using substitution of the form $t = s\cdot u(s)$ the equation is reduced to the $$u' = \frac{1}{2s} \implies u' = \frac{1}{2}\ln|C\cdot s|.$$ Using initial conditions constant C can be found using $y(5/2) = 1/4$, $s(5/2) = 4$, $t(4) = 0$, $u(4) = 0$, hence $0 = \ln|4C| \implies C = \pm \frac{1}{4}$.
  3. Putting it all together yields me $$y = \frac{1}{2}\left(x + \frac{3}{2}\right)\ln\left|\frac{x + 3/2}{4}\right| + 1/4$$ with two domains $I_1 = (-3/2, +\infty)$ and $I_2 = (-\infty, -3/2)$. But my textbook states that the answer is $$y = \frac{1}{2}\left(x + \frac{3}{2}\right)\ln\frac{x + 3/2}{4} + 1/4,~I = (-3/2, +\infty)$$

Hence the following two questions:

  1. What was the reason for removal of absolute value in the logarithm?
  2. Why is mine domain of validity wrong?
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1 Answer 1

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Since the initial condition is given for $x=\frac52$ we are assuming the solution for the domain $x>-\frac32\implies \frac{x + 3/2}{4}>0$.

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